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E = | F q | = k | qQ qr 2 | = k | Q | r 2 . size 12{E= { {F} over {q} } =k { { ital "qQ"} over { ital "qr" rSup { size 8{2} } } } =k { {Q} over {r rSup { size 8{2} } } } } {}

Since the test charge cancels, we see that

E = k | Q | r 2 . size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {}

The electric field is thus seen to depend only on the charge Q size 12{Q} {} and the distance r size 12{r} {} ; it is completely independent of the test charge q size 12{q} {} .

Calculating the electric field of a point charge

Calculate the strength and direction of the electric field E size 12{E} {} due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.

Strategy

We can find the electric field created by a point charge by using the equation E = kQ / r 2 size 12{E= { ital "kQ"} slash {r rSup { size 8{2} } } } {} .

Solution

Here Q = 2 . 00 × 10 9 size 12{Q=2 "." "00" times "10" rSup { size 8{ - 9} } } {} C and r = 5 . 00 × 10 3 size 12{r=5 "." "00" times "10" rSup { size 8{ - 3} } } {} m. Entering those values into the above equation gives

E = k Q r 2 = ( 8.99 × 10 9 N m 2 /C 2 ) × ( 2.00 × 10 9 C ) ( 5.00 × 10 3 m ) 2 = 7.19 × 10 5 N/C. alignl { stack { size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {} #= \( 9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} } \) times { { \( 2 "." "00" times "10" rSup { size 8{ - 9} } C \) } over { \( 5 "." "00" times "10" rSup { size 8{ - 3} } m \) rSup { size 8{2} } } } {} # =7 "." "20" times "10" rSup { size 8{5} } "N/C" {}} } {}

Discussion

This electric field strength is the same at any point 5.00 mm away from the charge Q size 12{Q} {} that creates the field. It is positive, meaning that it has a direction pointing away from the charge Q size 12{Q} {} .

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Calculating the force exerted on a point charge by an electric field

What force does the electric field found in the previous example exert on a point charge of –0.250 μ C ?

Strategy

Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field E = F / q size 12{E= {F} slash {q} } {} rearranged to F = q E size 12{F= ital "qE"} {} .

Solution

The magnitude of the force on a charge q = 0 . 250 μC size 12{q= - 0 "." "250""μC"} {} exerted by a field of strength E = 7 . 20 × 10 5 size 12{E=7 "." "20" times "10" rSup { size 8{5} } } {} N/C is thus,

F = qE = ( 0.250 × 10 –6 C ) ( 7.20 × 10 5 N/C ) = 0.180 N. alignl { stack { size 12{F= ital "qE"} {} #size 12{ {}= \( "-0" "." "250" times "10" rSup { size 8{"-6"} } `C \) \( 7 "." "20" times "10" rSup { size 8{5} } `"N/C" \) } {} # ="-0" "." "180"`N {}} } {}

Because q is negative, the force is directed opposite to the direction of the field.

Discussion

The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.

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Phet explorations: electric field of dreams

Play ball! Add charges to the Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

Electric Field of Dreams

Section summary

  • The electrostatic force field surrounding a charged object extends out into space in all directions.
  • The electrostatic force exerted by a point charge on a test charge at a distance r size 12{r} {} depends on the charge of both charges, as well as the distance between the two.
  • The electric field E size 12{E} {} is defined to be
    E = F q , size 12{E= { {F} over {q,} } } {}

    where F size 12{F} {} is the Coulomb or electrostatic force exerted on a small positive test charge q size 12{q} {} . E size 12{E} {} has units of N/C.

  • The magnitude of the electric field E size 12{E} {} created by a point charge Q size 12{Q} {} is
    E = k | Q | r 2 . size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {}

    where r size 12{r} {} is the distance from Q size 12{Q} {} . The electric field E size 12{E} {} is a vector and fields due to multiple charges add like vectors.

Conceptual questions

Why must the test charge q size 12{q} {} in the definition of the electric field be vanishingly small?

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Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field?

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Problem exercises

What is the magnitude and direction of an electric field that exerts a 2 . 00 × 10 - 5 N size 12{2 "." "00" times "10" rSup { size 8{5} } N} {} upward force on a –1.75 μ C charge?

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What is the magnitude and direction of the force exerted on a 3.50 μ C charge by a 250 N/C electric field that points due east?

8 . 75 × 10 4 size 12{8 "." "75" times "10" rSup { size 8{ - 4} } } {} N

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Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff).

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(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m?

(a) 6 . 94 × 10 8 C size 12{ {underline {6 "." "94" times "10" rSup { size 8{ - 8} } " C"}} } {}

(b) 6 . 25 N/C size 12{ {underline {6 "." "25"" N/C"}} } {}

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Calculate the initial (from rest) acceleration of a proton in a 5 . 00 × 10 6 N/C size 12{5 "." "00" times "10" rSup { size 8{6} } "N/C"} {} electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

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(a) Find the direction and magnitude of an electric field that exerts a 4 . 80 × 10 17 N size 12{4 "." "80" times "10" rSup { size 8{ - "17"} } N} {} westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?

(a) 300 N/C ( east ) size 12{ {underline {"300"" N/C " \( "eas"}} {underline {t \) }} } {}

(b) 4 . 80 × 10 17 N ( east ) size 12{ {underline {4 "." "80" times "10" rSup { size 8{ - "17"} } " N " \( "east" \) }} } {}

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Questions & Answers

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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Practice Key Terms 3

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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